X1、X2是方程X的平方+X-1=0的两根
1个回答
用对偶式解:
设4(X1)^2+10（X2)^3=A,则有4(X2)^2+10（X1)^3=B,
4(X1)^2+10（X2)^3=A...(1)
4(X2)^2+10（X1)^3=B.(2)
...(1)+.(2)得
4(X1^2+X2^2)+10（X1^3+X2^3)=A+B
4(X1+X2)^2-8X1X2+10(X1+X2)^3-30X1X2(X1+X2)=A+B
因X1+X2=-1,X1X2=-1
A+B=4+8-10-30=-28...(3)
...(1)+..-.(2)得
4(X1^2-X2^2)+10（X1^3-X2^3)=A-B.(4)
4(X1+X2)(X1-X2)+10(X1-X2)(X1^2+X1X2+X2^2)=A-B
4(X1+X2)(X1-X2)+10(X1-X2)[(X1+X2)^2-X1X2]=A-B
-4(X1-X2)+20(X1-X2)=A-B
16(X1-X2)=A-B
16√ [(X1+X2)^2-4X1X2]=A-B
A-B=16√5...(4)
.(3)+.(4)得
A=1/2(16√5-28)=8√5-14
所以4(X1)^2+10（X2)^3的值是8√5-14
相关问题
